Question: Find one value of $x$ that is a solution to the equation: $(3x-1)^2+12x-4=0$ $x=$
Solution: We could solve for $x$ by expanding $(3x-1)^2$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a shorter and more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form. Note that $12x-4=4({3x-1})$. This means that we can rewrite the equation as: $({3x-1})^2+4({3x-1})=0$ If we let ${p}={3x-1}$, we can see that this equation is in the form: ${p}^2+4{p}=0$ Let's solve this equation in terms of ${p}$ : $\begin{aligned}{p}^2+4{p}&=0\\\\ {p}({p}+4)&=0\\\\ {p}=0\ &\text{or} \ \ {p}=-4 \end{aligned}$ Since ${p}={3x-1}$, let's substitute this value back into our two solutions in order to solve for $x$ : ${3x-1}=0\ \ \ \text{or} \ \ \ {3x-1}=-4$ When we solve $3x-1=0$, we find that $x=\dfrac{1}{3}$. When we solve $3x-1=-4$, we find that $x=-1$. In conclusion, the two solutions of the equation $(3x-1)^2+12x-4=0$ are $x=\dfrac{1}{3}$ and $x=-1$. [Is there another way to solve for x?]